Question

What is `lim_(x->oo) (ln(x+1) - ln(x-1))/(1/x)`?

Answer 1

`2`

Explanation:

To prepare the product `x (log(x + 1) - log(x - 1))` for solution by l'Hôpital's rule (where `log` here is base `e`), rewrite it as:

`lim_(x->∞) (log(x + 1) - log(x - 1))/(1/x)`

Applying l'Hôpital's rule, we get that

`lim_(x->∞) ( d/( dx)(log(x + 1) - log(x - 1)))/( d/( dx)(1/x))`

`=lim_(x->∞) (1/(x + 1) - 1/(x - 1))/(-1/x^2)`

Multiplying the numerator and denominator by `(x+1)(x-1)x^2`

`= lim_(x->∞) (2 x^2)/((x - 1) (x + 1))`

`=2 lim_(x->∞) x^2/((x - 1) (x + 1))`

Let `u = x^2`. Then `lim_(x->∞) x^2/((x - 1) (x + 1)) = lim_(u->∞) u/(u - 1)`.

So, since `(x-1)(x+1)=(x^2-1)`, we have that

`= 2 lim_(u->∞) u/(u - 1)`

Divide the numerator and denominator by `u` gives:

`=2 lim_(u->∞) 1/(1 - 1/u)`

The expression `-1/u -> 0` as `u->∞`:

`2 lim_(u->∞) 1/(1 - 1/u)=2(1)=2`

Answer 2

`2`

Explanation:

Making `(x+1)/(x-1)=1+y` and solving for `x` we have `x = (1+2/y)` then

`((x+1)/(x-1))^x equiv (1+y)^(1/y+1) = (1+y)(1+y)^(2/y)`

and `lim_(x->oo) equiv lim_(y->0)` so finally

`lim_(x->oo) x(log(x+1)-log(x-1)) = log(lim_(x->oo)((x+1)/(x-1))^x) = log(lim_(y->0) (1+y)(1+y)^(2/y)) = log(e^2) = 2`